package algorithm.problems.greedy;

import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.Queue;

/**
 * Created by gouthamvidyapradhan on 27/06/2017.
 * <p>
 * There are n different online courses numbered from 1 to n. Each course has some duration(course length) t and closed on dth day. A course should be taken continuously for t days and must be finished before or on the dth day. You will start at the 1st day.
 * <p>
 * Given n online courses represented by pairs (t,d), your task is to find the maximal number of courses that can be taken.
 * <p>
 * Example:
 * Input: [[100, 200], [200, 1300], [1000, 1250], [2000, 3200]]
 * Output: 3
 * Explanation:
 * There're totally 4 courses, but you can take 3 courses at most:
 * First, take the 1st course, it costs 100 days so you will finish it on the 100th day, and ready to take the next course on the 101st day.
 * Second, take the 3rd course, it costs 1000 days so you will finish it on the 1100th day, and ready to take the next course on the 1101st day.
 * Third, take the 2nd course, it costs 200 days so you will finish it on the 1300th day.
 * The 4th course cannot be taken now, since you will finish it on the 3300th day, which exceeds the closed date.
 * <p>
 * Note:
 * The integer 1 <= d, t, n <= 10,000.
 * You can't take two courses simultaneously.
 * <p>
 * Solution: O(N log N)
 * 1. Sort the courses with earliest deadline time (Greedy sort)
 * 2. Maintain a max-heap of course duration.
 * 3. Iterate through each course and increment the total time by current course time and include this in the
 * max-heap created in step 2.
 * 4. If the total time exceeds the current course deadline then, remove the course with max duration from max-heap
 * inorder to accommodate the new course.
 */
public class CourseScheduleIII {
    public static void main(String[] args) throws Exception {
        int[][] course = {{5, 5}, {2, 6}, {4, 6}};
        System.out.println(new CourseScheduleIII().scheduleCourse(course));
    }

    public int scheduleCourse(int[][] courses) {
        Arrays.sort(courses, (a, b) -> a[1] - b[1]);
        Queue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
        int time = 0;
        for (int[] course : courses) {
            time += course[0];
            pq.add(course[0]);
            if (time > course[1])
                time -= pq.poll();
        }
        return pq.size();
    }
}
